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Homework Statement
Given a rod of length 'l' placed upright on a smooth floor.A slight disturbance causes it to fall.Now required is the velocity with which the COM falls down when the rod makes angle θ with horizontal.
The Attempt at a Solution
now let the middle portion of the rod be A,and the lower portion B.Using conservation of angular momentum,we get
0=(ml^2/12)ω-m*Va*l/2*cosθ
By work energy theorem,
mgl/2*(1-sinθ)=m(Va)^2/2+ml^2/12*ω^2 (ω about point B)
What is the mistake I'm making here?